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Question Number 85766 by jagoll last updated on 24/Mar/20
$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{1}−\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2y}\right) \\ $$
Commented by jagoll last updated on 24/Mar/20
![u = x+2y (du/dx) = 1+2(dy/dx) ((2dy)/dx) = (du/dx)−1 ⇒ (dy/dx) = (1/2)[ (du/dx)−1] ⇒ (1/2)(du/dx)−(1/2) = 1−sin u (du/dx) = 3−2sin u ⇒ (du/(3−2sin u)) = dx ∫ (du/(3−2sin u)) = x+c](Q85769.png)
$$\mathrm{u}\:=\:\mathrm{x}+\mathrm{2y} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{1}+\mathrm{2}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\frac{\mathrm{2dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{du}}{\mathrm{dx}}−\mathrm{1}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{du}}{\mathrm{dx}}−\mathrm{1}\right] \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{du}}{\mathrm{dx}}−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1}−\mathrm{sin}\:\mathrm{u} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{3}−\mathrm{2sin}\:\mathrm{u}\:\Rightarrow\:\frac{\mathrm{du}}{\mathrm{3}−\mathrm{2sin}\:\mathrm{u}}\:=\:\mathrm{dx} \\ $$$$\int\:\frac{\mathrm{du}}{\mathrm{3}−\mathrm{2sin}\:\mathrm{u}}\:=\:\mathrm{x}+\mathrm{c}\: \\ $$$$ \\ $$
Commented by john santu last updated on 25/Mar/20
![∫ (du/(3−2sin u)) = ∫ ((2dt)/((1+t^2 )(3−(4/(1+t^2 ))))) [ u = tan (t/2) ] ∫ ((2 dt)/(3t^2 −1)) = ∫ (dt/((√3) t−1)) −∫ (dt/((√3) t+1)) = (1/(√3)) ln ∣(((√3) t−1)/((√3) t +1))∣ = (1/((√3) )) ln ∣((2(√3) tan^(−1) (u)−1)/(2(√3) tan^(−1) (u)+1))∣ ∴ (1/(√3)) ln ∣((2(√3) tan^(−1) (x+2y)−1)/(2(√3) tan^(−1) (x+2y)+1))∣ = x+c](Q85805.png)
$$\int\:\frac{{du}}{\mathrm{3}−\mathrm{2sin}\:{u}}\:=\:\int\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{3}−\frac{\mathrm{4}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\left[\:{u}\:=\:\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\right] \\ $$$$\int\:\frac{\mathrm{2}\:{dt}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\int\:\frac{{dt}}{\sqrt{\mathrm{3}}\:{t}−\mathrm{1}}\:−\int\:\frac{{dt}}{\sqrt{\mathrm{3}}\:{t}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{3}}\:{t}−\mathrm{1}}{\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}\mid\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\:}\:\mathrm{ln}\:\mid\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left({u}\right)−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left({u}\right)+\mathrm{1}}\mid\: \\ $$$$\therefore\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{ln}\:\mid\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left({x}+\mathrm{2}{y}\right)−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left({x}+\mathrm{2}{y}\right)+\mathrm{1}}\mid\:=\:{x}+{c} \\ $$$$ \\ $$